# Prove that the operator induced by $T$ on the quotient space $V/\ker[T-5I]$ has all eigenvalues $=0$

The question is as follows : A linear operator T on a complex vector space V has a characteristic polynomial $x^3(x-5)^2$ and minimal polynomial $x^2(x-5)^2$ .

Now, i need to prove that the operator induced by T on the quotient space $V/\ker[T-5I]$ has all eigen values = $0$.

$Attempt$: The matrix from the given information can be deduced as follows :

\begin{pmatrix} 0 &1 &0 &0 &0\\ 0 &0 &0 &0 &0\\ 0 &0 &0 &0 &0\\ 0 &0 &0 &5 &0\\ 0 &0 &0 &0 &5 \end{pmatrix}

$$W = \ker(T-5I) = \operatorname{span} [(0,0,0,1,0)^T, (0,0,0,0,1)^T ].$$

The basis of the quotient is given by: $[ W+(1,0,0,0,0)^T, W+(0,1,0,0,0)^T , W+(0,0,1,0,0)^T]$.

Now, I don't understand specifically much by the phrase operator induced by T on the quotient space spanned by the above basis , but , i think it means T acting on a vector in the quotient space i.e. $T [ \alpha_1 (W + [1,0,0,0,0]^T ) + \alpha_2 (W + [0,1,0,0,0]^T + \alpha_3 (W + [0,0,1,0,0]^T ]$

\begin{align} & = T [W + \alpha_1 [1,0,0,0,0]^T + \alpha_2 [0,1,0,0,0]^T + \alpha_3 [0,0,1,0,0]^T ] \\ & = Tw + \alpha_1 .0 + \alpha_2 [1,0,0,0,0]^T + \alpha_3 .0 \end{align} $w$ is a vector which belongs to $W$

$$= T [ \alpha_4 [(0,0,0,1,0)^T + \alpha_5 (0,0,0,0,1)^T ] = 5\alpha_4 [(0,0,0,1,0)^T + 5\alpha_5 (0,0,0,0,1)^T$$

$$=(0,0,0,5\alpha_4, 5\alpha_5)^T$$

and the eigenvalues don't come as $0$ as desired in the question statement. Where am i going wrong in understanding the question? Thanks.

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Your way of writing subscripts and some of your other ways of writing MathJax code are bizarre. Please look at my edits. You shouldn't keep alternating in and out of MathJax so many times. –  Michael Hardy Dec 5 '13 at 19:54
Alright. Will take care of it. –  Wanderer Dec 5 '13 at 20:10

The quotient vector space has dimension $3$, and is spanned by the classes of the first three basis vectors. This you have computed above. In other words, the linear map on $W$ with respect to this basis has the matrix $$\begin{pmatrix} 0 & 1 & 0 \cr 0 & 0 & 0\cr 0 & 0 & 0 \end{pmatrix}.$$ This matrix has only $0$ as an eigenvalue, with multiplicity $3$.